< k2t+1+ 2k2/t2. Set t= logk/2, we obtain cr(G)<9k2/log2k. 2
4.4 Proof of Theorem 4.1.4
The idea and sketch of the construction.
In the descriprion we use weights on the edges of the graph. If we sub-stitute each weighted edge by an appropriate number of parallel paths, say,
each of length two, we can obtain an unweighted simple graph whose ratio of the pair-crossing and odd-crossing numbers is arbitrarily close to that of the weighted construction.
First of all, take a “frame”F, which is a cycle K with very heavy edges, together with a vertex V connected to all vertices of the cycle, also with very heavy edges. In the optimal drawings the edges ofF do not participate in any crossing, and we can assume that V is drawn outside the cycle K.
Therefore, all additional edges and vertices of the graph will be insideK. We have four further vertices, each connected to three different vertices of the frame-cycle K. These three edges have weights 1, 1, wrespectively, with some 1< w <2. Each one of these four vertices, together with the adjacent three edges, and the frame F, is called acomponent of the construction.
If we take anytwo of the components, it is easy to see how to draw them optimally, both in the odd-crossing and pair-crossing sense. See Figure 4.6.
The point is that if we take all four components, we can still draw them such that each of the six pairs are drawn optimally, in the odd-crossing sense. See Figure 4.7. On the other hand, it is easy to see that it is impossible to draw all six pairs optimally in the pair-crossing sense, some pairs will not have their best drawing. See Figure 4.8. Note that we did not indicate vertex V of the frame.
We get the best result with w= √5+12 . Actually, we will see that among any three components there is a pair which is not drawn optimally in the pair-crossing sense. So, we could take the union of just three components, but that gives a weaker bound.
2
Figure 4.6: (a) Component A (b), (c) Optimal drawings of the pairs (A, B) and (A, C), resp.
2
Figure 4.7: (a) Optimal drawing ofG in the odd-crossing sense (b), (c) The pairs (A, B) and (A, C) resp. from the same drawing.
2
Figure 4.8: (a), (b) Cases 1 and 2 of Lemma 2, resp., optimal drawings of G in the pair-crossing sense (c) Case 3, not optimal drawing.
Proof of Theorem 4.1.4.
Aweighted graphGis a graph with positive weights on its edges. For any edge e let w(e) denote its weight. For any fixed drawing G of G, the pair-crossing valuepair-cr(G) =Pw(e)w(e′) where the sum goes over all cross-ing pairs of edges e, e′. The odd-crossing value odd-cr(G) = Pw(e)w(e′) where the sum goes over all pairs of edgese, e′ that cross an odd number of times.
The pair-crossing number (resp. odd-crossing number) is the minimum of the pair-crossing value (resp. odd-crossing value) over all drawings. That
is,
pair-cr(G) = min
over all drawings
X
for all crossing pairs of edgese, e′
w(e)w(e′),
odd-cr(G) = min
over all drawings
X
for all pairs of edgese, e′
that cross an odd number of times
w(e)w(e′).
Theorem 4.4.1. There exists a weighted graphGwithpair-cr(G) = (3√25−
5
2)·odd-cr(G).
Proof of Theorem 4.4.1. First we define the weighted graph G. Take nine vertices, A1, B3, A2, C1, D3, C2, B1, A3, B2, D1, C3, D2 which form cycle K in this order. Vertex V is connected to all of the nine vertices ofK.
These vertices and edges form the “frame”F. All edges ofF have extremely large weights, therefore, they do not participate in any crossing in an optimal drawing. We can assume without loss of generality that V is drawn outside the cycle K, so all further edges and vertices of G will be inside K.
There are four more vertices, A0, B0, C0, D0, and forX =A, B, C, D, X0
is connected to X1, X2, and X3. The weight w(X0X1) = w(X0X2) = 1 and w(X0X3) = w = √5+12 . Graph X is a subgraph of G, induced by the frame and X0. See Figure 4.4.1. Finally, for any X, Y = A, B, C, D, X 6= Y, let pair-cr(X, Y) =pair-cr(X∪Y), and odd-cr(X, Y) =odd-cr(X∪Y).
First we find all these crossing numbers. Moreover, we also find out the second smallest pair-crossing values.
Start withA∪C. Since the path A1B3A2 is not intersected by any edge in an optimal drawing, we can contract it to one vertex, without changing the pair-crossing number, so now A1 = A2. Consider the edges e1 = A1A0
and e2 =A2A0. Now they connect the same vertices. Suppose that they do not go parallel in an optimal drawing. Let w∗(e1) (resp. v∗(e2)) be the sum of the weights of the edges crossing e1 (resp. e2) and assume without loss of generality that w∗(e1)≤w∗(e2). Then draw e2 parallel with e1, the drawing obtained is at least as good as the original drawing was, so it is optimal as well. Therefore, we can assume without loss of generality that e1 and e2 go parallel in an optimal drawing, so we can substitute them by one edge of weight 2. Similarly, we can contract the path C1D3C2 and substitute the edges C1C0 and C2C0 by one edge of weight 2. Now we have a very simple
graph, whose pair-crossing number is immediate, we have two pathsC1C0C3
andA1A0A3, which have to cross each other, and on both paths one edge has weight w the other one has weight 2. Since w < 2, in the optimal drawing the edgesA0A3 andC0C3 will cross each other and no other edges cross so we havepair-cr(A, C) = w2. Moreover, it is also clear that the second smallest pair-crossing value is 2w.
The same argument holds for odd-cr(A, C), moreover, by symmetry, we can argue exactly the same way for the pairs (A, D), (B, C), and (B, D).
Now we determine pair-cr(A, B) and the second smallest pair-crossing value. The edges a1 = A0A1, a2 = A0A2, a3 = A0A3 divide the interior of F into three regions R1, R2 and R3. Number them in such a way that for i = 1,2,3, ai is outside Ri. Once we place B0 into one of these regions, it is clear how to draw the edges b1 = B0BB1, b2 = B0BB2, b3 = B0BB3 to get the best of the possible drawings. If B0 is in R1 or in R2, we get the pair-crossing value 2w, but if we place B0 in R3, then we get 2. Again, the same argument holds for odd-cr(A, B), and by symmetry, the situation is the same with the pair (C, D).
Lemma 4.4.2.
odd-cr(G) = 4w2+ 4.
Proof of Lemma 4.4.2. We have odd-cr(G)≥odd-cr(A, B) +odd-cr (A, C) +odd-cr(A, D) +odd-cr(B, C) +odd-cr(B, D) +odd-cr(C, D)
= 4w2+4, and there is a drawing (see Fig. 4.7) with exactly this odd-crossing value. 2
Lemma 4.4.3.
pair-cr(G) = 4w2+ 4w.
Proof of Lemma 4.4.3. The argument, except for the exact calculation, should be clear from the figures. While we have a drawing which is optimal for all six pairs in the odd-crossing sense (see Fig. 4.7), in the pair-crossing sense some of the pairs will not be optimal, they have to take at least the second smallest pair-crossing value. We start with an observation that in any triple at least one pair is not optimal. Then we will distinguish three cases.
Take a drawing G of G. Suppose that we have a drawing G of G where the pairs (A, C) and (A, D) are drawn optimally, that is, pair-cr(A,C) = pair-cr(A,D) = w2. Recall that the edges a1 = A0A1, a2 = A0A2, a3 = A0A3divide the interior ofF into three regionsR1,R2 andR3. It follows from
the above argument that C0 ∈R1, D0 ∈R2. But then the pair (C, D) is not drawn optimally, that is, pair-cr(C,D) > 2, so we have pair-cr(C,D) ≥ 2w. In other words, it is impossible that all three pairs (A, C), (A, D), (C, D) are drawn optimally at the same time. By symmetry, this observation holds for any triple of A, B, C, D.
We have to distinguish three cases.
Case 1. Neither (A, B), nor (C, D) are drawn optimally. In this case, pair-cr(A,B)>2 so by the above argument we have pair-cr(A,B)≥2w, and similarly pair-cr(C,D)≥2w. For all other pairs we have pair-crossing value at least w2, therefore,pair-cr(G) =pair-cr(A,B) +pair-cr(A,C) + pair-cr(A,D)+pair-cr(B,C)+pair-cr(B,D)+pair-cr(C,D)≥4w2+4w.
Case 2. (A, B) is drawn optimally, (C, D) is not. Since (A, B) is drawn optimally, one of the pairs (A, C) and (B, C) and one of the pairs (A, D) and (B, D) is not drawn optimally so we have pair-cr(A,C) +pair-cr(B,C)≥ w2+2wand analogouslypair-cr(A,D)+pair-cr(B,D)≥w2+2wtherefore, pair-cr(G) = pair-cr(A,B) +pair-cr(A,C) +pair-cr(A,D) +pair-cr (B,C) +pair-cr(B,D) +pair-cr(C,D) ≥ 2w2+ 6w+ 2 = 4w2+ 4w. The last equality can be verified by solving the quadratic equation.
Case 3. Both (A, B) and (C, D) are drawn optimally. If none of the other four pairs is optimal, then we have pair-cr(G) = pair-cr(A,B) +pair-cr(A,C) +pair-cr(A,D) +pair-cr(B,C) +pair-cr(B,D) + pair-cr(C,D) ≥ 8w+ 4 = 4w2+ 4w. So we can assume that one of them, say (A, C) is drawn optimally, that is, pair-cr(A,C) = w2. Since in any triple we have at least one non-optimal pair, we have pair-cr(B,C) ≥ 2w and pair-cr(A,D)≥2w. We estimate pair-cr(B,D) now.
Again, the edges a1 = A0A1, a2 = A0A2, a3 = A0A3 of A divide the interior of F into three regions R1, R2 and R3 with Ri is the one to the opposite of ai. Similarly define the regions Q1, Q2, Q3 for C. Since (A, C) is drawn optimally, R3 and Q3 are disjoint. Since (A, B) is drawn optimally, B0 ∈R3, and since (C, D) is also drawn optimally, D0 ∈Q3. See Figure 4.8.
Now it is not hard to see that the edge D0D1 either crosses A0A1, A0A2, and B0B3, or B0B1, B0B2, and A0A3. The same holds for the edge D0D1, so pair-cr(A,D) + pair-cr(B,D) ≥ 2w+ 4. So we have pair-cr(G) = pair-cr(A,B) +pair-cr(A,C) +pair-cr(A,D) +pair-cr(B,C) +pair-cr (B,D) +pair-cr(C,D)≥w2+ 4w+ 8>4w2+ 4w. This concludes the proof of Lemma 4.4.3. 2.
Now we have
odd-cr(G)
pair-cr(G) = 4w2+ 4
4w2+ 4w = −5 2 + 3√
5 2 , and Theorem 4.4.1 follows immediately. 2
Return to the proof of Theorem 4.1.4. Let ε > 0 an arbitrary small number. Letpand q be positive integers with the property thatw(1 +10ε)>
p
q > w(1− 10ε). Let Gε be the following graph. In the weighted graph G of Lemma 4.4.3, (i) substitute each edge e = XY of weight 1 with q paths between X and Y, each of length 2, (ii) substitute each edge e = XY of weightwwithppaths betweenX andY, each of length 2, and (iii) substitute each edge e=XY of the frame F with a huge number of paths between X and Y, each of length 2. Then
odd-cr(Gε)
pair-cr(Gε) < odd-cr(G)
pair-cr(G)(1 +ε)< −5 2 + 3√
5 2 +ε.